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2 years ago

Why numbers are written as powers?

Mathematics

1 answer:

pishuonlain [190]2 years ago

3 0

Answer:

Answer below

Step-by-step explanation:

Large numbers like 100 billion are cumbersome to write out in ordinary form (100,000,000,000) and so scientists use a shorthand notation called powers of ten. ... The exponent tells you how many times ten is to be multiplied by iteslf to equal the number you wish to write. For example, 100 can be written as 10x10 = 102.

Hope this helps!

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Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

3 years ago

Y=x+3over 8 proportional or nonproportional

Inessa [10]

Answer:

nonproportional

Step-by-step explanation:

x + 3

------- is definitely nonproportional. The reason for this is that the graph (a

8 straight line) does not pass through the

origin; it's offset by 3/8 unit up from the

origin.

6 0

2 years ago

Lines a and b are parallel. Find the measure of each angle. Angle 8 is 113 degrees

pentagon [3]

Answer:

angles 1 4 and 5 are all equal to 113 and angles 2 3 6 and 7 equal 67

3 0

3 years ago

Devonte used the change of base formula to approximate log Subscript 8 Baseline 25. Which expression did Devonte use?.

scZoUnD [109]

Answer:

I think I did this. I think the answer choices are:

Log StartFraction 8 Over 25 EndFraction

StartFraction log 8 Over log 25 EndFraction

log StartFraction 25 Over 8 EndFraction

----> StartFraction log 25 Over log 8 EndFraction

So the answer is D

Step-by-step explanation:

6 0

2 years ago

I need help please and thank you

marta [7]

So what u need to do is

46+x=180 and u will get the answer

5 0

3 years ago